Collection of Solved Problems in Physics

Pressure, Volume and Temperature of a Compressed Gas

Task number: 2180

• Notation

  p1 = 0.1 MPa = 105 Pa	initial pressure of the gas
  T1 = 293 K	initial temperature of the gas
  V1 = 830 l = 0.83 m3	initial volume of the gas
  W = 166.8 kJ = 166.8·103 J	work performed during compression of the gas
  κ = 1.25	Poisson's ratio
  p2 = ?	pressure of the gas after compression
  V2 = ?	volume of the gas after compression
  T2 = ?	temperature of the gas after compression

• Hint 1

  As it is said in the task assignment, the process takes place in a thermally isolated system. What does it mean?

• Solution of Hint 1

  If the system is thermally isolated, the process which it is undergoing is adiabatic.

• Hint 2: Determining Volume V₂

  Final volume of the gas V₂ can be determined from the formula for the performed work.

  Think about what this formula looks like when pressure of the gas is not constant (but is the function of volume).

• Formula for Work W

  Since the pressure p in this task is not constant, yet it is the function of volume, i.e. p(V), integration will be necessary to calculate the work. Thus, we use the following formula for the performed work

  \[W = -\int\limits_{V_1}^{V_2}p(V)\, \text{d}V,\]

  where V₁ stands for initial volume and V₂ for final volume of the gas.

  First, we need to express pressure p as a function of volume V, and then we could integrate.

• Expressing Pressure

  To express pressure p as a function of volume V, use the Poisson's equation for adiabatic processes.

• Poisson's Equation

  Poisson's equation applies to adiabatic processes as follows:

  \[pV^{\kappa}=const,\]

  where p stands for pressure, V for volume and κ for the Poisson's ratio.

• Hint 3: Final Pressure p₂

  To calculate the final pressure p₂ of the gas, use the above-mentioned Poisson's equation.

• Hint 4: Final Temperature T₂

  To calculate the final temperature T₂ of the gas, use the ideal gas law.

• Analysis

  To determine the final volume of the gas, we use the formula for the performed work. Since pressure is a function of volume, we will have to use the integral form of this formula. Thus, we express pressure from the Poisson's equation first and then substitute it in the integral for the obtained function. Once integrated, we can express the final volume.

  After that we use the Poisson's equation again to determine the final pressure of the gas.

  Final temperature of the gas after compression can be determined from the ideal gas law.

• Solution

  Since pressure of the gas p pressure during compression is not constant, the following formula for the performed work holds true

  \[W = -\int\limits_{V_1}^{V_2}p \, \text{d}V,\]

  where V₁ and V₂ are initial and final volume of the gas respectively.

  To express pressure p as a function of volume V, we use the Poisson's equation for adiabatic processes:

  \[p_1V_{1}^{\kappa} = pV^{\kappa}, \]

  where κ is the Poisson's ratio. After expressing pressure p from this equation, we obtain

  \[p = \frac{p_1V_1^{\kappa}}{V^{\kappa}}.\]

  Now we substitute pressure in the formula for the work

  \[W = -\int\limits_{V_1}^{V_2}p \, \text{d}V = -\int\limits_{V_1}^{V_2}\frac{p_1V_1^{\kappa}}{V^{\kappa}} \, \text{d}V =\]

  factor out the constants and pull them out of the integral

  \[= -p_1V_1^{\kappa} \int\limits_{V_1}^{V_2}\frac{1}{V^{\kappa}} \, \text{d}V =\]

  integrate and substitute for the limits

  \[= -p_1V_1^{\kappa}\frac{1}{-\kappa + 1}\left[V^{-\kappa + 1}\right]_{V_1}^{V_2}= \frac{p_1V_1^{\kappa}}{\kappa-1 }\left(V_2^{-\kappa+1} - V_1^{-\kappa+1}\right).\]

  Therefore, work performed during the gas compression is

  \[W= \frac{p_1V_1^{\kappa}}{\kappa-1}\left(V_2^{1-\kappa} - V_1^{1-\kappa}\right) = \frac{p_1V_1}{\kappa-1}\left[\left(\frac{V_2}{V_1}\right)^{1-\kappa} - 1\right] .\]

   

  Therefrom we can express the final volume of the gas V₂:

  \[\left(\frac{V_2}{V_1}\right)^{1-\kappa}= \frac{(\kappa-1)W} {p_1V_1}+1,\]

  and so:

  \[V_2=V_1 \sqrt[1-\kappa]{\frac{(\kappa-1)W}{p_1V_1}+1}.\]

  Then, using the Poisson's equation again

  \[p_1V_1^{\kappa}=p_2V_2^{\kappa} \]

  we can easily determine the resultant pressure of the gas p₂:

  \[p_2=\frac{p_1V_1^{\kappa}}{V_2^{\kappa}}.\]

  Finally, we use the following form of the ideal gas law

  \[\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2},\]

  from which we express the resultant temperature of the gas T₂ after compression:

  \[T_2=\frac{p_2V_2T_1}{p_1V_1}.\]

• Numerical Substitution

  \[V_2=V_1\sqrt[1-\kappa]{\frac{(\kappa-1)W}{p_1V_1}+1}\] \[V_2=0.83\cdot\sqrt[1-1.25]{\frac{(1.25-1)\cdot166.8\cdot{10^3}}{10^5\cdot{0.83}}+1}\,\mathrm{m^3}\dot{=}0.163\,\mathrm{m^3}\] \[p_2=\frac{p_1V_1^{\kappa}}{V_2^{\kappa}}=\frac{10^5\cdot{0.83^{1.25}}}{0.163^{1.25}}\,\mathrm{Pa}\dot{=}765000\,\mathrm{Pa}=765\,\mathrm{kPa}\] \[T_2=\frac{p_2V_2T_1}{p_1V_1}=\frac{765000\cdot{ 0.163}\cdot{293}}{10^5\cdot{ 0.83}}\,\mathrm{K}\dot{=}440\,\mathrm{K}\]

• Answer

  The resultant pressure, volume and temperature of the compressed gas are approximately 765 kPa, 0.163 m³ and 440 K respectively.